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Adiabatic Expansion of an Ideal Gas

Adiabatic Expansion of an Ideal Gas

A process carried in a vessel whose walls are perfectly insulated so that no heat can pass through them, is said to be adiabatic.

In adiabatic process there is no heat exchange between a system and surroundings, and q = 0.

According to the First law:

ΔE = q – w = 0 – w
ΔE = – w          ...(1)

Since the work is done at the expense of internal energy, the internal energy decreases and the temperature falls.

Consider 1 mole of an ideal gas at pressure P and a volume V. For an infinitesimal increase in volume dV at pressure P, the work done by the gas is –PdV. The internal energy decreases by dE.

According to equation (1):

dE = – PdV      ...(2)

By definition of molar heat capacity at constant volume:

dE = CvdT    ...(3)

From (2) and (3):

CvdT = – PdV

For an ideal gas:

P = RT/V

and hence:

Integrating between T1, T2 and V1, V2 and considering Cv to be constant:

Since R = Cp – Cv , this equation may be written as:

The ratio of Cp to Cv is often written as γ :

and equation (4) thus becomes:

Replacing – ve sign by inverting V2/V1 to V1/V2 and taking antilogarithms:

We can also eliminate the temperature by making use of the ideal gas relationship:

Equating the right-hand sides of equations (5) and (6):

Comparison between Isothermal and Adiabatic Expansions

Boyle’s law describes pressure-volume relations of an ideal gas under isothermal conditions (T, constant). This is similar to the relation derived for adiabatic expansion.

PV = constant (Boyle’s law)
PVγ = constant (Adiabatic expansion)

γ for an ideal monoatomic gas = 1.67.

The difference between the two processes is : in an isothermal process, temperature of a system remains constant while in an adiabatic process, temperature must change.


In an isothermal process heat is absorbed to make up for the work done by the gas in expansion and the temperature remains unchanged.

On the other hand, adiabatic expansion takes place at the expense of internal energy which decreases and the temperature falls. For the same reason, the curve for the adiabatic process is steeper than that for the isothermal process.

Work done In adiabatic Reversible Expansion

Step (1): Value of VdP from adiabatic equation

For an adiabatic process:

PVγ = constant

Differentiating it, we have:

γ PVγ – 1 dV + Vγ dp = 0

Dividing by V γ – 1, we get:

γ PdV + VdP = 0
VdP = – γ PdV              ...(1)

Step (2): Value of VdP from ideal gas equation

For 1 mole of an ideal gas:


Complete differentiation gives:

PdV + VdP = RdT
VdP = RdT – PdV            ...(2)

Step (3): Substitution

Substituting the value of VdP from (1) in (2) we get:

RdT – PdV = – γ PdV
RdT = P (1 – γ) dV

If there are n moles of a gas:

Step (4): Integration

Integrating from T1, V1 to T2, V2 with γ constant:

When T2 > T1, wmax is negative because 1 – γ is negative. This means that work is done on the

On the other hand, when T2 < T1, wmax is positive which means that work is done by the gas.

Solved Problem

Problem (1):Calculate w for the adiabatic reversible expansion of 2 moles of an ideal gas
at 273.2 K and 20 atm to a final pressure of 2 atm.

Cp  = 5R/2, mole–1deg–1
Cv  = 3R/2, mole–1deg–1
R = 8.314J mole–1deg–1

Step 1: To calculate the value of T2, the final temperature, using the equation:

(T2 / T1)γ = (P2 / P1)γ –1

Substituting the value of γ in (1):

(T2 / 273.2)5/3 = (2 / 20)

Solving it, we get:

T2 = 108.8 K

Step 2: To calculate maximum work under adiabatic conditions:

The work done under adiabatic conditions may be obtained by calculating decrease in internal

w = – ΔE = – nCv (T2 – T1)
= – 2 × 3 / 2 × 8.314 (108.8 – 273.2)
= 4100 J = 4.1 kJ

Problem (2): At 25ºC for the combustion of 1 mole of liquid benzene the heat of reaction at constant pressure is given by:

What would be the heat of reaction at constant volume?

We have

Reference: Essentials of Physical Chemistry /Arun Bahl, B.S Bahl and G.D. Tuli / multicolour edition.

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